From: matt@... (Matt Neuburg)
Date: 2012-09-13T12:50:21+09:00
Subject: Re: Breaking Down the Block

Think of it this way:

inject(memo) { | memo, enum | block sets memo } => memo

So memo has three stages:

1. Before the first iteration, memo is set to the value of the parameter
(in parentheses after "inject").

2. After each iteration, the result of the block becomes the new value
of memo.

3. When it's all over, the value of memo is the result.

So it's all about accumulating results into memo. (It is sometimes
called the accumulator.)

In your example, the accumulation involves adding the successive values
of enum to the accumulator. The successive values of enum are the values
of the array. So the result is the sum of the array. (It's the sum of
the array plus the initial value of the memo, but that initial value is
zero, so it really is just the sum.)

m.

incag neato <lists@ruby-forum.com> wrote:

> Can someone please explain in plain english how this block treats the
> given variables (and values in the array)? I am struggling to understand
> the relationship of 't' and 'n' with the array.
> 
> class Array
>   def sum
>     inject(0) {|t,n| t + n}
>   end
> end
> puts [1,2,3].sum
> 
> Is 't' effectively labeled as '0' (zero), given the assignment on the
> inject method?
> What "is" 't' exactly?


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